Nim.

The game goes like this: create any number of piles of objects with any number of objects in each pile. You could have a pile of 3 red objects, a second pile of four blue objects, and a third pile of five green objects, for example. (I’ve color-coded the objects to make the game easier to understand).

On your turn, remove any number of objects, with two rules: you must remove at least one object, and all the objects you remove must be from the same pile.

If you remove the last object, you win.

The game is solved. Convert the number of objects in each pile into a binary number. Count the total number of 1’s in each place across all piles (in the sample piles above, there’s a 11 pile, a 100 pile, and a 101 pile, leading to two 1’s in the fours place, one 1 in the two’s place, and two 1s in the one’s place). If there are an even number of 1s in each place, then it’s a “winning position,” with “winning position” defined as “if you end your turn like this and play right, you’re guaranteed to win.” If there are an odd number of 1s in any place, it’s a losing position, with “losing position” defined as “if you end your turn like this, your opponent has a chance to win.”

The most basic winning position is two piles with 1 in each pile. Converted to binary, there are two 1s in the one’s place, which is an even number. If you end your turn with this position, then your opponent must take 1 cube (since they’re in two different piles), and you can then take the other cube and win.

I understand **that** it’s solved, and I understand how to determine whether any position is a winning position or a losing position.

What I don’t get is why. Why base 2 instead of base 3? Why must it be in pairs?

I feel like I’m on the verge of understanding it, but I don’t quite get it. If someone can explain it narratively to me–that is, without using fancy symbology, but rather laying it out in layman’s terms–I’d be very grateful.